ID	a	m	p	Predecessor	e	ES	EF	LS	LF	SLACK
1
2	20	25	30		25	0	25	0	25	0
3	38	40	45	2	40.5	25	65.5	25	65.5	0
4	15	20	25	2	20	25	45	45.5	65.5	20.5
5	9	10	12	3,4	10.16667	65.5	75.66667	65.5	75.66667	0
6	12	15	18	5	15	75.66667	90.66667	193.3333	208.3333	117.6667
7	25	30	40	5	30.83333	75.66667	106.5	104	134.8333	28.33333
8	35	50	55	5	48.33333	75.66667	124	75.66667	124	0
9	10	10	15	8	10.83333	124	134.8333	124	134.8333	0
10	20	25	30	7,9	25	134.8333	159.8333	134.8333	159.8333	0
11	5	7	8	10	6.833333	159.8333	166.6667	166.5	173.3333	6.666667
12	10	14	15	10	13.5	159.8333	173.3333	159.8333	173.3333	0
13	30	35	40	11,12	35	173.3333	208.3333	173.3333	208.3333	0
14	1	1	1	6,13	1	208.3333	209.3333	208.3333	209.3333	0

Expected time of each activity, e = (a+4m+p)/6

The estimated completion time of the project is 209.33 rounded to 209 days

Explanation: The duration of the project has been determined by forward pass.

The ES and EF are calculated from the start to end activities of the network.

ES or Earliest Start Time = EF or Earliest Finish time of the predecessor of the corresponding activity

EF = ES+ Duration

if an activity has more than 1 predecessor than the maximum of the EF of the predecessors is considered as the ES for the corresponding activity.

Example: Activities 2 does not have any predecessors, hence can be started at 0 time. So the ES is 0.

Activity 3 has one predecessor which is 2. So, the ES = EF of 3 = 25

5 has two predecessors, 3 and 4.The ES of activity 5 = Max (EF of Activity 3 and 4)

The highest EF time is the duration of the project.

LS or Late Start time = LF or Late Finish Time - Duration of the corresponding project.

LF = LS of the successor

The LF and LS are calculated from the end to start the activity. Hence the activity with the highest EF becomes the end activity and the EF becomes the LF.

If an activity has more than one successor, the minimum of the LS values is considered for the LF.

Example: Activity 14 has the highest EF time. This means this activity finishes last. Hence the latest finish for the project is 209.33. Hence LF = 209.33

Activity 14 has two predecessors 6 and 13. Hence the LS of 14 becomes the LF of both 6 and 13

Activity 10 is the predecessor of two activities 11 and 12 So, the LF = Min (LS of 11, LS of 12)

Similarly, the ES, EF, LS, and LF of all other activities are calculated as shown in the above table.

Slack = LF - EF or LS - ES

Activities with Slack = 0 are the critical activities and the corresponding path is the critical path.

Answers:

1. The project estimated to be completed in 209.33 days

3 The critical path for this project is 2-3-5-8-9-10-12-13-14

4. 9 tasks are on the critical path

5. The buffer should be 7.8 days

explanation:

ID	a	m	p	Predecessor	e	variance
1
2	20	25	30		25	2.777778
3	38	40	45	2	40.5	1.361111
4	15	20	25	2	20	2.777778
5	9	10	12	3,4	10.16667	0.25
6	12	15	18	5	15	1
7	25	30	40	5	30.83333	6.25
8	35	50	55	5	48.33333	11.11111
9	10	10	15	8	10.83333	0.694444
10	20	25	30	7,9	25	2.777778
11	5	7	8	10	6.833333	0.25
12	10	14	15	10	13.5	0.694444
13	30	35	40	11,12	35	2.777778
14	1	1	1	6,13	1	0

variance of each activity = ((p-a)/6)^2

project variance = sum of variances of critical activities = 2.777778+1.361111+0.25+11.11111+0.694444+2.777778+0.694444+2.777778+0 = 22.44

z value for 95% confidence = 1.65

z = (x-e)/standard deviation

standard deviation = project variance = 22.44 = 4.73

1.65 = (x-209.33)/4.73

=> x = 1.65*4.73+209.33 = 217.13

The desired duration for the 95% confidence of project completion is 217.13

The buffer should be = 217.13-209.33 = 7.8 days

Answer part b

Part B

Table B1 shows the personnel that have been assigned to the project and their respective rates per hour. Table B2 shows their assignment to different project tasks.

5. Use the file from Part 1 and the information contained in Tables B1 and B2 to assign resources to the project schedule. Save a copy of this file as ZumaBaseLineB1.

6. Create a Memo that answers the following questions:

1. What is the total cost of the project?

2. Which if any of the resources are over-allocated?

3. Which activities involve over-allocated resources?

Resource $/hour Number Available Marketing Specialists $60 4 Design Engineers $90 4 Development Engineers $80 4 Industrial Engineers $70 4 Purchasing Agent $50 1 ID 2 Description Market Analysis Product Design Resources Marketing Specialist[400%] Design Engineers[400%] 3 4 Manufacturing Study Industrial Engineers[200%] Product Design Selection Design Engineers[300%] 5 6 7 Detailed Marketing Plan Manufacturing Process Detailed Product Design Marketing Specialist[400%] Industrial Engineers[400%] Design Engineers[400%] 8 9 Test Prototype Design Engineers[300%] 10 Finalized Product Design Marketing Specialist[200%] 11 Order Components 12 Order Production Equipt 13 Install Production Equipt Purchasing Agent[100%] Purchasing Agent[100%] Industrial Engineers[400%] ID Description 14 Celebrate Resources Marketing Specialist[400%];Design Engi- neers[400%];Development Engineers[400%];Industrial Engineers[400%];Purchasing Agent[100%] Resource $/hour Number Available Marketing Specialists $60 4 Design Engineers $90 4 Development Engineers $80 4 Industrial Engineers $70 4 Purchasing Agent $50 1 ID 2 Description Market Analysis Product Design Resources Marketing Specialist[400%] Design Engineers[400%] 3 4 Manufacturing Study Industrial Engineers[200%] Product Design Selection Design Engineers[300%] 5 6 7 Detailed Marketing Plan Manufacturing Process Detailed Product Design Marketing Specialist[400%] Industrial Engineers[400%] Design Engineers[400%] 8 9 Test Prototype Design Engineers[300%] 10 Finalized Product Design Marketing Specialist[200%] 11 Order Components 12 Order Production Equipt 13 Install Production Equipt Purchasing Agent[100%] Purchasing Agent[100%] Industrial Engineers[400%] ID Description 14 Celebrate Resources Marketing Specialist[400%];Design Engi- neers[400%];Development Engineers[400%];Industrial Engineers[400%];Purchasing Agent[100%]